MATH 290 – HOMEWORK 2 – SOLUTIONS
1. Prove that for any integers
a
and
b
,
a
+
b
is even if and only if either both
a
and
b
are
even or both
a
and
b
are odd.
Solution:
Let
a
and
b
be integers.
(=
⇒
) We will prove this direction by contrapositive. Suppose that it is not true
that either both
a
and
b
are even or both
a
and
b
are odd. This means that one
of
a
or
b
is odd and the other even. Let us assume that
a
is even and
b
is odd (the
proof is exactly the same if we assume the other way around). Then for some
integers
k
and
m
,
a
= 2
k
and
b
= 2
m
+1. Then
a
+
b
= 2
k
+2
m
+1 = 2(
k
+
m
)+1.
Since
k
+
m
is an integer,
a
+
b
is odd hence not even.
(
⇐
=) We will prove this by examining two cases.
Case 1.
Assume that both
a
and
b
are even. Then for some integers
k
and
m
,
a
= 2
k
and
b
= 2
m
. Then
a
+
b
= 2
k
+2
m
= 2(
k
+
m
). Since
k
+
m
is an integer,
a
+
b
is even.
Case 2.
Assume that both
a
and
b
are odd. Then for some integers
k
and
m
,
a
= 2
k
+ 1 and
b
= 2
m
+ 1. Then
a
+
b
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 Fall '08
 Alligood,K
 Advanced Math, Real Numbers, Integers, Negative and nonnegative numbers, positive real numbers

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